Originally Posted by

**Henryt999** First of all thanks to both of you for your help.

You are correct there should be an i there where ProveIt said.

Accutually I have to solve this in three different ways.

By simplifying a+bi

by rewritting it in polar form

By using (as tonio did) Eulers approach.

Also this has to be done without a calculator, (that probably means that the argument rational).

Anyway the issue is that I donīt understand what I do wrong with the polar form thing: I just took the arguments in the denominator that gave me:

$\displaystyle \frac{\pi}{4}+\frac{\pi}{3} = \frac{7\pi}{12}$

then the arguments in the nominator: $\displaystyle \frac{\pi}{2}+\frac{\pi}{6} = \frac{2\pi}{3}$

This is already a mess: in a fraction the upper part is called NUMERATOR and the lower one DENOMINATOR. Now, in the numerator you have $\displaystyle z_1=1+i\,,\,\,z_2=1+\sqrt{3}\Longrightarrow \arg z_1=\arctan(1\slash 1)=\pi\slash 4$ $\displaystyle arg(z_2)=\arctan(\sqrt{3})=\pi\slash 3$ , and in the denominator you have $\displaystyle z_3=3i\,,\,\,z_4=\sqrt{3}-i\Longrightarrow \arg(z_3)=\pi\slash 2\,,\,\,\arg(z_4)=-\pi\slash 6$.

Now you do the maths with the powers and you'll get $\displaystyle \pi\slash 4$ , indeed.

Tonio

and then substracted that gave me $\displaystyle \frac{7\pi}{12}-\frac{8\pi}{12} = -\frac{\pi}{12}$

but the correct answer is $\displaystyle \frac{\pi}{4}$

not sure what Iīm doing wrong here...